What mass of precipitate will form if 1.50 L of concentrated Pb(ClO3)2 is mixed with 0.750 L of 0.170 M NaI? Assume the reaction goes to completion.
To determine the mass of precipitate formed, we first need to write the balanced chemical equation for the reaction between Pb(ClO3)2 and NaI:
Pb(ClO3)2 + 2NaI -> 2NaClO3 + PbI2
From the balanced equation, we can see that the molar ratio between Pb(ClO3)2 and PbI2 is 1:1. This means that for every one mole of Pb(ClO3)2, we get one mole of PbI2.
Next, let's calculate the number of moles of Pb(ClO3)2 and NaI involved in the reaction:
To calculate the number of moles of Pb(ClO3)2:
- Concentration of Pb(ClO3)2 is not given, so we assume it is 100% concentrated, meaning it is 100% Pb(ClO3)2.
- Molar mass of Pb(ClO3)2 = 207.2 g/mol (lead (Pb) is 207.2 g/mol, and each ClO3 group is 35.45 + 3 * 16.00 = 125.45 g/mol)
- Volume of Pb(ClO3)2 = 1.50 L
Using the equation:
moles of Pb(ClO3)2 = (concentration * volume) / molar mass
moles of Pb(ClO3)2 = (1 * 1.50) / 207.2
Next, let's calculate the number of moles of NaI:
- Concentration of NaI = 0.170 M
- Volume of NaI = 0.750 L
Using the equation:
moles of NaI = concentration * volume
moles of NaI = 0.170 * 0.750
Now, we compare the moles of Pb(ClO3)2 and NaI to determine the limiting reagent. The limiting reagent is the one that is completely consumed, thus determining the maximum amount of product formed.
- If the moles of Pb(ClO3)2 are higher, it is the limiting reagent.
- If the moles of NaI are higher, it is the limiting reagent.
Once we identify the limiting reagent, we know that the number of moles of PbI2 formed is equal to the number of moles of the limiting reagent.
Now, using the obtained moles of PbI2, we can calculate the mass of the precipitate:
- Molar mass of PbI2 = 461.0 g/mol (lead (Pb) is 207.2 g/mol, and each I atom is 126.9 g/mol)
- Mass of PbI2 formed = moles of PbI2 * molar mass of PbI2
By following this process, you can calculate the mass of precipitate formed when 1.50 L of concentrated Pb(ClO3)2 is mixed with 0.750 L of 0.170 M NaI.